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受欢迎的三角函数 >

sin(x)=1+cos^2(x)

解答

sin (x) = 1 + cos^{2} (x)

解答

x = \frac{π}{2} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

sin (x) = 1 + cos^{2} (x)

两边减去

1 + cos^{2} (x)

sin (x) - 1 - cos^{2} (x) = 0

使用三角恒等式改写

- 1 - cos^{2} (x) + sin (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 1 - (1 - sin^{2} (x)) + sin (x)

化简

- 1 - (1 - sin^{2} (x)) + sin (x) : sin^{2} (x) + sin (x) - 2

- 1 - (1 - sin^{2} (x)) + sin (x)

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

- (1 - sin^{2} (x))

打开括号

= - (1) - (- sin^{2} (x))

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + sin^{2} (x)

= - 1 - 1 + sin^{2} (x) + sin (x)

数字相减:

- 1 - 1 = - 2

= sin^{2} (x) + sin (x) - 2

= sin^{2} (x) + sin (x) - 2

- 2 + sin (x) + sin^{2} (x) = 0

用替代法求解

- 2 + sin (x) + sin^{2} (x) = 0

令

: sin (x) = u

- 2 + u + u^{2} = 0

- 2 + u + u^{2} = 0 : u = 1, u = - 2

- 2 + u + u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

u^{2} + u - 2 = 0

使用求根公式求解

u^{2} + u - 2 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

1^{2} - 4 \cdot 1 \cdot (- 2)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 3}{2 \cdot 1}, u_{2} = \frac{- 1 - 3}{2 \cdot 1}

u = \frac{- 1 + 3}{2 \cdot 1} : 1

\frac{- 1 + 3}{2 \cdot 1}

数字相加/相减:

- 1 + 3 = 2

= \frac{2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- 1 - 3}{2 \cdot 1} : - 2

\frac{- 1 - 3}{2 \cdot 1}

数字相减:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

二次方程组的解是:

u = 1, u = - 2

u = sin (x)

代回

sin (x) = 1, sin (x) = - 2

sin (x) = 1, sin (x) = - 2

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - 2 :

无解

sin (x) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{π}{2} + 2 πn

作图

流行的例子

cos^2(x)+4cos(x)+1=0

cos^{2} (x) + 4 cos (x) + 1 = 0

arcsin(x)= pi/4

arcsin (x) = \frac{π}{4}

arcsin(x)= pi/3

arcsin (x) = \frac{π}{3}

sin^{2} (x) = 4

6-sin(θ)=cos(2θ)

6 - sin (θ) = cos (2 θ)